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Q.
A satellite in a circular orbit of radius $R$ has a period of $4\, h$. Another satellite with orbital radius $3\,R$ around the same planet will have a period (in hours)
AFMCAFMC 2008
Solution:
According to Kepler's third law
$T^{2} \propto R^{3}$
$\Rightarrow \frac{T_{2}}{T_{1}}=\left(\frac{R_{2}}{R_{1}}\right)^{3 / 2}$
$\therefore \frac{T_{2}}{T_{1}}=\left(\frac{3 R}{R}\right)^{3 / 2}$
$\Rightarrow \frac{T_{2}}{T_{1}}=\sqrt{27}$
$\Rightarrow T_{2}=\sqrt{27 T_{1}}=\sqrt{27} \times 4$
$=4 \sqrt{27 h}$